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Let $G = HD$ be a group with $H \leq G$ and $D \unlhd G$. Suppose that $d\in D$ and denote $K = \langle H, H^{d}\rangle$ and $B = D \cap K$. I need to show that $K = HB$

Since $H \leq K$ and $G= HD$, it follows that $B \unlhd K$, so that $HB \leq K$

I can't seem to show that $K\leq HB$. Clearly $H \leq HB$. If I show that $H^{d} \leq HB$ I would be done

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    There is something wrong. From your definitions, we have $HB = H$, and clearly it is not necessarily true that $K \le H$. I wonder if you meant $B = D \cap K$? – 2017-02-23
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    If you define $B = D \cap K$, then an element of $H^d$ has the form $d^{-1}hd$ and since $G=HD$, we get $d^{-1}hd = h'd'$ and then $d' \in D \cap K$, so $H^d \le HB$ as required. – 2017-02-23
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    @DerekHolt Yes thank you for picking up that typo precisely. $B = D \cap K$ – 2017-02-23

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